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Soliton in a Well. Dynamics and Tunneling.
V. Fleurov
Raymond and Beverly Sackler Faculty of Exact Sciences,
School of Physics and Astronomy,
Tel-Aviv University, Tel-Aviv 69978 Israel.
A. Soffer
Department of Mathematics,
Rutgers University, New Brunswick, NJ 08903,USA
May 18, 2013
Abstract
We derive the leading order radiation through tunneling of an oscillating soliton in a well. We use the hydrodynamic formulation with
a rigorous control of the errors for finite times.
1
Introduction
The dynamics of nonlinear dispersive systems pose a great challenge to mathematics and physics. A large number of possible phenomena, which include
the formation of solitons or other coherent structures, the blow up solutions,
multichannel scattering, and nonlinear tunneling, makes the development of
a ”general theory” practically impossible. We then look at specific systems.
Here we will consider the motion of a soliton, described by the Nonlinear
Schrödinger (NLS) equation, moving inside a one-dimensional well, which is
however truncated so as to allow tunneling.
This problem was considered in great detail by [1, 2, 3]. They showed
that in the semiclassical parameter regime a soliton, which is close to the
bottom of the potential well, will stop moving as time goes to infinity. The
1
process of stopping is due to radiation out of the well via tunneling. When
viewed formally, the coupling of the bound state to the continuum shows
up at 1/ω order of perturbation expansion. Here ω = 2π/T where T is the
oscillation period. It was shown in [1, 2] (see also[4]) that the corresponding
time dependent resonance theory [5, 23] requires 1/ω orders of iteration.
Our approach to this problem is different. We only solve the problem
up to a finite time, though very long compared to the oscillation period T .
In fact our results apply to times of the order t0 = 1/(ωε) where ε is the
oscillation amplitude. Previous work on the finite time semiclassical soliton
was developed in detail and for a general potential in [6]. In that paper only
the classical trajectory for the soliton is derived, but not the dissipation due
to quantum tunneling. We prove using the hydrodynamic formalism of the
quantum tunneling [7, 8, 9, 10] that the radiation outside the well is outgoing
and find its leading order behavior. The fact that the radiation is purely
outgoing is then used to prove that it implies the monotonic decay in time
of the internal energy of the soliton and of its oscillation amplitude. These
estimates hold up to times of order t0 . So the extension to an arbitrary time
is possible, since it implies that in the time interval 0 ≤ t ≤ t0 the amplitude
ε(t) goes monotonically down.
The control of the errors in the hydrodynamic formulation is done here for
the first time. The key difficulty is to estimate the error due to the quantum
potential. The latter is defined as
Q(|ψ|) = −
∂x2 |ψ|
|ψ|
for the wave function ψ. Let ψ be written as
ψ = S(x − a(t))(1 + χ(x, t))eıθ
where S(x) is the static soliton profile at the bottom of the trap. The control of the error in Q(|ψ|) is reduced to estimating the multiplicative type
perturbation χ.
A further consequence of our analysis is that with a particular choice of
the potential at the transition region, we observe a significant suppression of
tunneling, which stabilizes the soliton oscillation dynamics for a long time.
2
2
Soliton in the closed potential well V (x)
We consider in this section motion of a soliton in a closed well from which
tunneling is impossible. The dynamics in this case is governed by the GrossPitaevskii (GP) equation (Nonlinear Schrödinger equation - NLS)
i∂t ψ = −∂x2 ψ + λ|ψ|2 ψ + V (x)ψ(x, t)
(1)
(the Planck constant ~ = 1 and the mass m = 1/2). Here we assume that we
know the real function S(x) of the soliton sitting at the minimum at x = 0
of the trap potential V (x), which solves the equation
−ES(x) = −∂x2 S(x) + λS 3 (x) + V (x)S(x)
(2)
Then the approximate solution corresponding to a moving soliton is looked
for in the form
ψ(x, t) = S(x − X(t))e+i(Et+Φ)
(3)
with a real phase Φ. X(t) defines the position of the soliton at a given time t.
Substituting this solution into Eq. (1) and separating the real and imaginary
terms we get two equations
[
]
∂t Φ + (∂x Φ)2 + W (x, t) S(x − X(t)) = 0
(4)
where W (x, t) = V (x) − V (x − X(t)) and
[∂t X(t) − 2∂x Φ] ∂x S(x − X(t)) = ∂x2 ΦS(x − X(t))
(5)
Assuming now that the soliton S(x − X(t)) is very narrow even function
centered around X(t) we may integrate Eq. (4) over x and approximately
obtain the equation
∂t Φ(X, t) + (∂x Φ(X, t))2 + V (X) = 0
(6)
which is a Hamilton-Jacoby equation describing the motion of a particle with
the mass 1/2 at the center X of the soliton. W (X) = V (X) − V (0) should
have been written in Eq. (6), however the constant V (0) in the potential
may be always omitted. The phase Φ(X) plays the role of mechanical action
and the velocity of the soliton motion is
Ẋ = 2∂X Φ(X).
3
Hence for x close to X the Ansatz
Φ(x, t) ≈ Ẋ(t)x/2 + F (t)
(7)
may be applied. Here the function F (t) depends only on t. As a result, Eq.
(5) becomes an identity. Substituting Eq. (7) into (4) we get in the same
approximation that
t
1
F (t) = ẊX −
2
0
∫t [
]
1
2
Ẋ(s) − V (X(s)) ds
4
(8)
0
which really has the form of a mechanical action with a boundary term which
does not affect equation of motion for X(t).
Correspondingly the total energy of the soliton within the trap becomes
1
Etot = −E − V (0) + Ẋ(t)2 + V (X(t)) = const
4
The corrections to the above approximate derivation are due to the finite
width of the soliton,
∫
−2
βe = dxx2 S(x).
(9)
which is supposed to be small compared to the characteristic scale of the
potential.
It is important to emphasize that in the case of harmonic potential V (x) =
2 2
ω x /4 the approximate solution (3) and (7) for a soliton moving in the trap
becomes exact with
X(t) = C0 cos 2ωt
where C0 is the oscillation amplitude and
1
F (t) = − ωC02 sin 2ωt.
8
This can be verified by a direct substitution into Eqs. (4) and (5). As for
the total energy it is obviously
1
Etot = −E − V (0) + C02 ω 2 .
4
For other types of nontrivial exact solutions, see e.g. [20, 17].
4
3
Estimates of the corrections
3.1
”Time dependent case”
In order to estimate the precision of the applied procedure we return to Eq.
(1) which describes a soliton oscillating within a trap and radiation due to
tunneling. Hence the chemical potential E(t), (Energy), and the shape of
function S(x) vary with time t.
The general solution will be looked for using the following Ansatz (see
[1, 2, 3, 21, 22] for related constructions):
ψ = SE (x − X(t), t)(1 + χ)eiθ ,
(10)
where
ψt=0 = S(x − Xm , 0).
with some initial value Xm . SE (x, t) is the solution of the nonlinear soliton
equation
−E(t)S(x, t) = (−∂x2 + V (x) + F (S))S.
for a given value of E(t), which may vary adiabatically slow with time. We
distinguish here the rapid time dependence due to oscillations of the soliton
within the trap described by the shift X(t) and slow variation of the shape
of the function ψt=0 due to exponentially weak tunneling.
Substituting (10) into (1) (the subscript E is suppressed for the sake of
brevity) we get
iṠ(1 + χ)eiθ + iS χ̇eiθ − S(1 + χ)θ̇eiθ =
[ 2
]
−∂x ψ + V ψ + [F (S(1 + χ)) − F (S) + F (S)]ψ
[
= eiθ (−S ′′ + V S + F (S)S) (1 + χ)
(
)
S′ ′
′′
2
∗
+ S −χ − 2 χ + λS (χ + χ )
S
]
+ λS 3 (χ2 + 2|χ|2 + |χ|2 χ)
[
Se′
+ eiθ S(1 + χ) (θ′ )2 − iθ − 2iθ′
Se
′′
]
(11)
where Se = S(1 + χ). Here · stands for the time derivative and ′ stands for
the x derivative.
5
Now, we use the Ansatz
∫
θ(x, t) =
t
E(s)ds + γ(t) + φ(x, t),
0
such that
2∂x θ = vθ = vφ = 2∂x φ.
Then, the Schrödinger equation becomes
ıṠ(1 + χ) + ıS χ̇ − S(1 + χ)(φ̇ + γ̇) = (V − Vt )(1 + χ)S
(
)
+ S −χ′′ − 2(S ′ /S)χ′ + λS 2 (χ + χ∗ )
+ λS 3 (χ2 + 2|χ|2 + |χ|2 χ)
e θ′ /2).
+ S(1 + χ)vθ2 /4 − ı(Se′ vθ + Sv
(12)
The complexified equation for the vector
( )
χ
K=
χ∗
can then be written as
ıS
where
−1
(
Ṡ + ıK̇ = HK + O(bK, b) + N L(K) + O
)
S′
′
vθ + vθ .
S
(13)
( 2
)
−∂x + λS 2 − 2(S ′ /S)∂x
λS 2
H :=
,
−λS 2
∂x2 − λS 2 + 2(S ′ /S)∂x
S stands for the vector with components S, −S; b stands for time dependent,
space localized functions, which (are expected to) have a zero limit at infinite
time. N L stands for nonlinear terms in K.
Orthogonality Conditions Let HD := S(H − Eσz )S −1 then
( )
S
HD
= 0,
−S
(
)
∂E S
2
HD
= 0.
∂E S
6
(14)
(15)
∗
Similar identities hold for the adjoint operator HD
. The corresponding eigenfunctions are obtained from the previous ones, by applying the Pauli matrix
σz .
HD has also two generalized eigenfunctions, which are small perturbations
of the vectors
(
)
(
)
∂x S
xS
η1 =
,
η2 =
,
∂x S
−xS
with eigenvalues close to zero, provided the soliton is sufficiently narrow. See
[1, 2].
(
)
∂E S
ξ1 =
.
∂E S
We now impose the main orthogonality conditions, which make the perturbation χ, orthogonal to the soliton S.
Re⟨S 2 , χ⟩ = 0,
′
Im⟨χ, SSE ⟩ = 0,
Im⟨Sx′ , Sχ⟩ = 0.
where ⟨φ, ψ⟩ denotes the usual L2 scalar product of φ and ψ. The first two
conditions determine the equations of Ė, γ̇. The other conditions determine
the boost and translation of the Soliton at time t.
For example, multiplying (13) by S, taking the scalar product with the
vector σ3 ξ1 , and using the orthogonality condition, using also the equation
∂t S(x − X(t), E(t)) = −∂x S(x − X(t))Ẋ(t) + ∂E S Ė.
for Ṡ, integration gives
∫
2γ̇ S∂E Sdx = ⟨σ3 ξ1 , SO(bK, b) + SN L(K) + O(Svθ )⟩.
Going back to equation (12), if we choose
∫ t
1
θh (t) =
E(s)ds + xẊ(t) + q(x, t),
2
0
7
(16)
(17)
and
ψ = S(1 + χ)eıθh = S|1 + χ|eıq+ıθh = S|1 + χ|eıϕ ,
we derive
vθh = 2∂x θh = Ẋ(t) = vϕ − vq .
Finally, we derive the following equations for χ:
[
0=
−∂x [V (x) − V (x − X) − xẌ/2]
+
]
S
2
2
−1
[ (Imχ̇ + Hχ Reχ + 2λS Reχ + O(χ ))] + S̃ ∂E S ĖImχ .
S̃
From this we derive
(18)
−∂t Imχ = Hχ Reχ + (2λS 2 + γ̇)Reχ + O(χ2 )+
S̃ −1 ∂E S ĖImχ + [V (x) − V (x − X) − xẌ/2 + γ̇].
4
(19)
Hydrodynamic representation
In order to find the tunneling flux from an open potential Vext (x) with a
soliton oscillating within the well we apply the procedure similar to that
used in our previous papers[7, 8, 9, 10]. First the GP equation (1) for a
complex function ψ = |ψ| exp{iϕ} can be represented as two hydrodynamic
equations,[18, 19],
∂t ρ(x, t) + ∂x [ρ(x, t)v(x, t)] = 0.
(20)
and
[
]
√
∂x2 ρ(x, t)
1
1
∂t v(x, t) + v(x, t)∂x v(x, t) = −∂x Vext (x) − √
+ λρ(x, t) (21)
2
2
ρ(x, t)
for two real functions: the density ρ(x, t) = |ψ|2 and the velocity v = vϕ =
2∂x ϕ.
8
Theorem 4.1. Hydrodynamic Equations. Let ψ satisfy the NLS equation,
with initial data of finite energy; assume, moreover that the solution exists
and is bounded. Then, if for all (x, t) ∈ [IX , IT ], ψ ̸= 0, then the hydrodynamic equations (20) and (21) for ρ, v have a unique solution identified with
the corresponding NLS, via the relations ρ = |ψ|2 , ψ = |ψ|eıϕ , v = 2∂x ϕ.
For a proof see [11]. The general problem of approximating the Schrödinger
type equations with kinetic type or fluid equations, has been studied extensively. It is important in understanding semiclassical methods[12, 13], large
particle system limits and more[14, 15, 16].
In order to solve the problem we choose |ψ| = S0 (x − X(t)) obtained
near the bottom of the potential Fig. 1. That is S0 (x) is an exact solution
stationary at the bottom of the potential well. We assume that the initial
condition is that the solution at time t = 0 is given by the soliton profile
S(x − Xm ) with Xm small. Here, it is assumed that the potential have a
local minimum at zero. (The function S0 is taken to solve the nonlinear
soliton equation (2).) Then
∂t v(x, t) + v(x, t)∂x v(x, t) = −2∂x (U1 + U2 )
(22)
U1 = [V (x) − V (x − X(t))]
(23)
U2 = F(S(1 + χ)) − F(S) + Q + V (x − X(t)) + F(S)
(24)
with
and
where Q = − ∂x|ψ||ψ| is the quantum potential (QP) and F stands for the
nonlinear term. We also use the ansatz
2
ψ = S(x − X(t))(1 + χ(x, t))eiθ
(25)
where θ = θh + q, with θh solving the harmonic problem with the frequency
ω.
The derivative
∂x [Vh (x) − Vh (x − X(t))] = ω 2 X(t)/2
for some constant ω, so that (m = 1/2),
∂t2 X(t) = −ω 2 X(t)
9
and
v(x, t) = Ẋ(t) + w(x, t).
We use that Q(S) = −∂x2 S/S, and
−
∂x2 S
+ V (x − X(t)) + F(S) = −E = const
S
and ψ = S(1 + χ)eiθ , as well as the estimate of Q(ψ) − Q(S) to be presented
in the next subsection. Then the equation for w becomes
∂t w(x, t) + Ẋ(t)∂x w(x, t) + w(x, t)∂x w(x, t) = −2∂x W1 (x, t)
(26)
with
W1 (x, t) = V (x) − Vh (x) + Vh (x − X(t)) − V (x − X(t))+
F(S(1 + χ)) − F (S) + Q(ψ) − Q(S),
where the correction w(x, t) appears due to tunneling and corresponds to the
escape of the matter from the trap. Alternatively, the equation for v is
∂t v(x, t) + v(x, t)∂x v(x, t) =
−2∂x [W (x, t)(1 + χ) + O(Sχ) + Q(ψ) − Q(S) + O(χ2 )].
(27)
with
W (x, t) = V (x) − V (x − X(t))
The solution of this equation can be found from the solution of the corresponding classical mechanical system. After approximating the quantum
potential as Q(ψ) ∼ Q(S) + O(χ), we obtain the equation
∂t v(x, t) + v(x, t)∂x v(x, t) =
−2∂x [W (x, t)(1 + χ) − L(Reχ) + O(Sχ2 + χ∂x χ)].
(28)
for the velocity v.
L(Reχ) ≡ Hχ Reχ + 2λS 2 Reχ + O(ĖSReχ).
See Section 4 for the estimate of Q(ψ) − Q(S).
To find the solution for the classical system, let us assume first that the
potential looks like piecewise harmonic: It is harmonic well, around the origin, up to distance of order 1/ω, then a transition region, inverted harmonic,
10
on an interval of order 1, and then goes down as the same harmonic potential,
all the way to zero.

1 2 2
ω x,
|x| ≤ ω1 ,

4


1
1
2
2
2
3

|x ∓ ω1 | ≤ ωδ ,
 − 4 ω (x ∓ ω ) ± ω O(x )
1 2
2
ω (x − ω2 )2 ,
≥ x > 1+δ
,
V (x) =
4
ω
ω

1 2
2 2
2
1+δ

ω (x + ω ) ,
−ω ≤ x < − ω ,


4

0
|x| > ω2
with 0 < δ ≪ 1.
Theorem 4.2. Burgers’ Equation.
Consider the Burgers equation defined in (28), and with zero initial data.
The solution of this equation, without the second order terms corrections, in
the transition region ω2 ≥ x > 1+δ
, is given by
ω
[
]
1
v(x, t) = αϵ2 ω 2 t − (t2 − βϵ−2 ω −2 (x − (1 + δ)/ω))1/2
2
(29)
for t2 ≫ βϵ−2 ω −2 (x − (1 + δ)/ω)) . Moreover, for x in the transition region,
and (x, t) not satisfying the above inequality, the velocity v(x, t) is zero.
Remark The above result shows that tunneling is suppressed for this
choice of the potential barrier, for times of order β 1/2 ϵ−1 ω −1/2 .
Proof. In this case, the potential field acting on the classical particle in the
transition region is approximately given by
V (x) − V (x − X)
= V ′ (x)X − (1/2)V ′′ (x)X 2 + O(X 3 )
∼ ω 2 xX − α(x)ω 2 X 2 + C(t) + O(X 3 )
(30)
with a positive sn (x)α, since in the transition region, the third derivative of
V is positive, of order ω 2 on an interval of a size O(1). Here, sn (x) denotes
the sign of x. So, the force is outgoing,
F ∼ −ω 2 X + α0 ω 2 X 2 = −ω 2 X + [(1/2)α − cos(2ωt)]ϵ2 ω 2 .
(31)
where we choose α(x) ∼ α0 x + β0 , and since X(t) ≡ ϵ sin ωt with a small
oscillation amplitude ϵ.
11
So, the classical particle goes out of the transition region, with the velocity
v0 ∼ ϵ2 (a/2)t0 + dẊ, the exit time is t0 . Hence
1 2 22
ϵ ω t0 ∼ 1
2
(32)
so that
1
(33)
ωϵ
(The third derivative is of the order ω 2 only on an interval of size O(1)). To
the leading order, d = 1. So, outside the transition region, and AFTER time
t′ , the acceleration is zero and the velocity is therefore given by v(x0 , t′ ), and
up to Ẋ is positive, of the order ϵ2 ω 2 t′ , for (x − 1/ω) > 0 . So, to find the
solution for the velocity at (x, t), we get
t0 ∼
x − x0 = v(x0 , t′ )(t − t′ )
for some t′ ≤ t. Since the acceleration is constant, when it is nonzero, by our
choice of the potential, we get that:
1
x − x0 = αϵ2 ω 2 t′ (t − t′ ).
2
Solving for t′ , gives TWO possible solutions. The solution of the Burgers
equation is the one which minimizes the action. The action is proportional
to (t′ )3 , and therefore is minimized by the smaller choice of t′ . This gives the
result of the theorem.
4.1
Corrections to the quantum potential
Here we show that assuming small χ, linear corrections to the quantum
potential
∂ 2 |ψ|
Q=− x
|ψ|
make important contribution. We will prove that:
Theorem 4.3. Suppose that |χ| << 1, and the solution of the NLS is given
as before by our Ansatz , in terms of the Soliton S,, and the correction χ.
Then the Quantum potential Q is given by
Q=−
∂x2 S
+ Hχ Reχ + O(χ∂x χ) + ReN L(χ, χ∗ )
S
12
∂x2 S
− ∂t Im χ − [V (x) − V (x − X(t))]
S
+L(Reχ) + ReN L(χ, χ∗ ) + O(χ∂x χ) + O(Imχ∂t Reχ).
Q=−
with
Hχ = −∂x2 −
(34)
2∂x S
∂x .
S
Proof. Using
|ψ| = S|1 + χ|
we write
Q=−
∂x2 S 2∂x S∂x |1 + χ| ∂x2 |1 + χ|
−
−
S
S|1 + χ|
|1 + χ|
(35)
We may now calculate the derivatives
∂x |1 + χ| =
∂x Reχ + (∂x Reχ)Reχ + (∂x Imχ)Imχ
|1 + χ|
and
∂x2 |1 + χ| =
∂x [∂x Reχ + (∂x Reχ)Reχ + (∂x Imχ)Imχ] (∂x Reχ)2
−
+ O(χ3 )
|1 + χ|
|1 + χ|2
Q=−
∂x2 S 2∂x S 1
−
[∂x Reχ(1 + Reχ) + (∂x Imχ)Imχ)] −
S
S |1 + χ|
]
1 [ 2
∂x Reχ(1 + Reχ) + (∂x Reχ)2 + (∂x Imχ)2 + Imχ∂x2 Imχ
|1 + χ|
+O(χ3 )
since
(∂x Reχ)2 −
(∂x Reχ)2
= O(χ3 ).
|1 + χ|
Then the force coming from Q is
( 2
)
∂x S 2∂x S
2
−∂x Q = −∂x −
−
(∂x Reχ) − (∂x Reχ)
S
S
)]
[(
)(
1
2∂x S
2
−∂x
−1
−
(∂x Reχ) − (∂x Reχ)
|1 + χ|
S
13
(36)
−
[
]}
1 {
−∂x Reχ∂x2 Reχ + (∂x Imχ)2 + O(χ2 ∂x χ) + ∂x O(∂x2 (Imχ)2 )
|1 + χ|
= O(χ∂x χ) + O(∂t (χ∂x χ))
Next we eliminate ∂x2 Reχ, ∂x2 Imχ in the above equations, using the equations
−∂t Imχ = Hχ Reχ + O(S 2 χ + χ2 ) + G(W (x, X(t))).
(37)
∂t Reχ = Hχ Imχ + O(S 2 χ + χ2 ).
(38)
Therefore the quantum potential becomes in the leading order
Q=−
∂x2 S 2∂x S
−
∂x Reχ − ∂x2 Reχ + O(χ∂x χ)
S
S
+O((∂x Imχ)2 + Imχ∂t Reχ).
(39)
Now we use equation
i∂t χ = Hχ χ + L(χ) + N L(χ, χ∗ ) + G(W (x, X(t)))
where L and N L stand for linear and nonlinear terms in χ. It allows us to
write
(
)
2∂x S
2
−∂t Imχ = − ∂x +
∂x Reχ+Re(L(Sχ)+N L(χ, χ∗ ))+G(W (x, X(t))),
S
which implies the statement of the Theorem.
In order to proceed further on we will need some relations between the
correction q to the soliton phase due to χ and the phase η of the latter. They
are connected by equation
1 + |χ|eiη = |1 + χ|eiq .
(40)
We will also need the derivative
1
[∂x |χ|2 +2 cos η∂x |χ|+2|χ| sin η∂x η]. (41)
∂x |1+χ| = √
2
2 1 + |χ | + 2|χ| cos η
Equation (40) for complex functions is equivalent to
1 + |χ| cos η = |1 + χ| cos q,
|χ| sin η = |1 + χ| sin q.
14
(42)
(43)
5
Hydrodynamic Modulation Equations
In this section we combine the estimate for the velocity being outgoing in the
outer well, together with the equations for χ, Ė, R, to prove that the soliton
stabilizes. Recall the ansatz for the solution, in terms of the soliton S and
the phase θ:
ψ = S(1 + χ)eıθ := S|1 + χ|eı[θ+q] .
Since we have shown that the velocity is outgoing, modulo the oscillating
part, we have that (choosing θ = θh ),
sign(x)∂x q ∼ ω 2 ϵ2 t > 0, t ≤ t0 ∼
1
.
ωϵ
Theorem 5.1. Soliton Energy Decay.
′
Let the solution of NLS be such that |χ| + |χ | << 1, ∂x q > 0. Assume,
moreover that the soliton solution S is stable. Then
Ė ≤ −c(ω)Reχ∂x q(x0 )
for some x0 at the transition region, and c(ω) is a constant of order O(e−bx0 ),
and b is a positive constant related to the decay rate of the soliton (in x) at
large distances.
Proof. We have
∫ x
∂t
|ψ|2 dx = −j(ψ(x)) = −2S 2 |1 + χ|2 (∂x θ + ∂x q)
−∞
= −2S 2 (∂x θ +∂x q)−2S 2 (Reχ)Ẋ(t)−S 2 |χ|2 (Ẋ(t)+vq )−4S 2 (Reχ)∂x q. (44)
Using the equation
|ψ|2 = S 2 + 2S 2 Reχ + S 2 |χ|2 ,
(45)
and combining equations (16), ( 44), (45) and 2∂x θh = Ẋ we derive that
∫ x
∫ x
2
∂t
|R| + 2Ė(t)
S∂E S = −S 2 [|1 + χ|2 − 1]Ẋ(t)
−∞
−∞
[
∫
− 2S ∂x q + 2S (2Reχ + |χ| )∂x q + ∂t
2
2
x
15
]
2S Reχ .
2
2
−∞
(46)
As a result of the orthogonality conditions, the modulation equation for
R = Sχ reads
∫ x
∂t
|R|2 dx = −j(R) + O(N L(R)) + Im {R∗ S[F (|ψ|) − F (S)]}
∫
= ∂t
with
−∞
∫
+∞
∞
|R| dx − ∂t
2
−∞
∫
|R| dx = −∂t
x
∫
+∞
2
+∞
|S| dx − ∂t
|R|2 dx. (47)
2
−∞
x
j(R) = S 2 j(χ) = −ı(R∗ ∂x R − R∂x R∗ ).
(48)
Next, we use that
1 + χ = S −1 |ψ|eıq = |1 + χ|eıq .
So that Reχ = |χ| cos η = |1 + χ| cos q − 1, Imχ = |χ| sin η = |1 + χ| sin q.
Using the assumption that χ, ∂x χ, and q, are all small in the relevant region
of space and time, we arrive at
∂x Imχ = (∂x q)|1 + χ| cos q + (∂x |1 + χ|) sin q =
∂x q [|1 + χ| cos q] + 2∂x Reχ|χ| sin η + O(χ2 )
= ∂x q + |χ| cos η∂x q + 2|χ|∂x q sin2 η(2 cos η + sin η) + O(χ2 )
= ∂x q(1 + O(Reχ)) + O(q 2 )∂x q + O(q)(∂x Reχ)(1 + O(χ)).
(49)
We now compute the derivative of Reχ with respect to time, to leading
order. For this we use the equality
∫ x
∫ ∞
2
−2∂t
S Reχdx = 2∂t
S 2 Reχdx,
(50)
−∞
x
which follows from the orthogonality (S 2 , Reχ). It is obvious that the integral
in the right hand side of this equality is exponentially small. Then
∫ ∞
+2∂t
S 2 Reχdx =
x
∫ ∞
2
S 2 ⟨[(−∂x2 − 2S −1 ∂x S∂x )Imχ]⟩dx +
∫x ∞
∫ ∞
3
+
S O(χ)dx + 4(−Ẋ(t))
∂x SSReχdx
x
x
∫ ∞
+ 4Ė
S∂E SReχdx,
(51)
x
16
by the equation ∂t Reχ = H(Imχ) + O(S 2 χ). Integration by parts of the
operator −∂x2 − 2S −1 ∂x S∂x yields
∫ ∞
S 2 Reχdx =
2∂t
x
∫
∞
2
2S ∂x Imχ +
∫
S 2 ⟨[+O(χS 2 )⟩]dx
x
∫
∞
+4Ė
∞
S∂E SReχdx + 4(−Ẋ(t))
S∂x SReχdx.
(52)
x
x
The last term can be represented in the form
∫ ∞
4(−Ẋ(t))
S∂x SReχdx
∫
= −2Ẋ(t)
x
∞
⟨∂x (S 2 Reχ) − S 2 (∂x Reχ)⟩dx
x
∫
2
∞
= +2Ẋ(t)S Reχ + 2Ẋ(t)
S 2 (∂x Reχ)dx.
(53)
x
Starting from the differential equation for R (see [1]) we derive that the
quantity
∫ ∞
∂t
|R|2 = 2S 2 |χ|2 ∂x q + O(χ3 S 3 + · · · )
(54)
x
is of the order of |χ| S(x)2 ∂x q .
Putting all in Eq. (46) we get
∫ +∞
∫
2
2
2
3 3
−2S |χ| ∂x q − ∂t
|S| dx + O(χ S + · · · ) + 2Ė
2
−∞
2
x
−∞
S∂E Sdx
= −2S ∂x q − 4S 2 (Reχ)∂x q
∫ ∞
2
2
− 2S |χ| ∂x (q) + 2Ẋ(t)
S 2 (∂x Reχ)dx + 2S 2 ∂x q
x
+ 2S |χ| cos η∂x q + 2S |χ|∂x q sin2 η(2 cos η + sin η)
∫ ∞
∫ ∞
2
2 m
+ 2
S ⟨[O(χ S )⟩]dx + 4Ė
S∂E SReχdx.
2
2
x
∫
−∂t
x
∫
+∞
|S| dx = −2
−∞
∫
+∞
2
−∞
S∂E S Ėdx + 2Ẋ(t)
17
+∞
−∞
S∂x Sdx
(55)
∫
= −2Ė
+∞
S∂E Sdx.
−∞
(56)
As a result we arrive at the equation
[ ∫ x
∫ +∞
∫
Ė(t) 2
S∂E Sdx −
S∂E Sdx − 4
−∞
[
∫
Ė −2
−∞
S∂E SReχdx
x
∫
x
−∞
]
∞
S∂E S − 4
∞
]
S∂E SReχdx dx
x
= −2S 2 (Reχ)∂x q + 2S 2 |χ|∂x q sin2 η(2 cos η + sin η)
∫ ∞
2
2
2
2
−2S |χ| ∂x (θ + q) + 2S |χ| ∂x q + 2Ẋ(t)
S 2 (∂x Reχ)dx
x
∫ ∞
+2
S 2 O(χ2 S m )dx.
(57)
x
for Ė(t). We also note that for sin η = ± √25 , cos η = ∓ √15 , the sin2 η term is
zero to leading order. See Theorem 5.3.
Next, we find the leading order term on the right hand side of equation
(57). We compute first
∂x Reχ = cos η∂x |χ| − |χ| sin η∂x η
= cos η sin ηvq (1 + O(χ)) + sin η(sin η + cos η + O(χ))vq
= vq sin η [cos η + (sin η + cos η + O(χ))]
√
Now, at the trapping values of η, that is, when sin η = ±2/ 5, the above
trigonometric expression vanishes identically. When cos η = 0, the integral
multiplying the Ẋ(t) factor is seen to be zero, by direct integration by parts,
since now Reχ = |χ| cos η = 0, by assumption. Hence
∂x Reχ = vq O(χ).
Since Ẋ(t) is of order ϵω ≪ 1, we conclude that the Ẋ(t) term is of higher
order correction to the term −2S 2 (Reχ)∂x q. All other terms are higher order
in χ.
Finally we get our principal result, (at trapped cos η = ± √15 ):
(∫
)−1
x
Ė(t) =
−∞
S∂E S
S 2 Reχ∂x q + O(S 2 [χ2 v + Ẋ(Reχ)2 ]).
18
(58)
At cos η ∼ 0, we get that Reχ is replaced by |χ|, in the above formula. If
cos η is positive at some time, it will stay there, and flow either to zero or to
+ √15 . The same is true if it is negative, not too far from zero. This is because
the equation for the derivative of Reχ with respect to time, is positive to
leading order, for the region where ∂x q is positive. Using the soliton stability
condition, it follows that Ė(t) is negative, to the leading order in χ.
We now need to show that χ stays small in the relevant neighborhood of
space and time, that is, around the transition region, including the range of
the classical trajectories, that contribute to the solution of the Euler equation,
up to the time desired. The equation for χ can be written as
(
)
ıχ̇ = −∂x2 − 2S −1 ∂x S∂x χ + 2SReχ + O(χ2 )
+O(W ) + O(Ėχ) + O(γ̇χ).
(59)
This is done in the next section.
5.1
Multiplicative Perturbation Bounds
Bounding χ .
In this section we derive bounds on |χ| and |χ|′ , using the estimates for
q ′ and the (bootstrap) assumption of small χ. Recall that
1 + |χ| cos η = |1 + χ| cos q
|χ| sin η = |1 + χ| sin q
In particular, it follows that
sin q = q + O(χ3 ) = O(χ).
We have the following bounds on |χ|′ :
Proposition 5.2. For 0 < |χ| << 1 ,
|χ|′ ≤ c|q ′ |.
|η ′ | ≤
c|q ′ |
|χ|
19
(60)
(61)
Proof. For |χ| ≪ 1 we may write that (see Section 7)
q′
|1 + χ|[+ sin q(1 + cot η) − |χ| cos η] + O(|χ|q ′ )
|χ|
|χ|′ =
(62)
and
|χ| sin η = |1 + χ| sin q.
(63)
|χ|′ ≤ c|q ′ || sin η|
(64)
|χ|′ = q ′ sin η + O(|χ|q ′ ).
(65)
Hence
and,
Now we use
∫
∫
b
∂x |χ| dx = |χ(b)| − |χ(a)| = 2
2
2
2
a
b
|χ||χ|′ dx
(66)
a
and from equation (93) of Section 7
q′
q′
[cos η + sin η + O(|χ|)] ≤ c
η =−
|χ|
|χ|
′
for |χ| ≥ ε|q ′ |. To see that the higher order terms do not blow up at the
limits when either sin η or cos η approach zero, we use the asymptotic bound
on |χ|′ , in equations (87) and (88) of Section 7: In the limit when cos η
approaches zero, we get that
η ′ |χ|2 sin η ≈ q ′ − |χ|′ + O(q ′ |χ|) = O(q ′ |χ|).
Hence in this limit we get
|η ′ | ≤ c|q ′ |/|χ|
. Similarly, in the limit sin η near zero, we get:
η ′ |χ| ≈ q ′ + |χ|′ ≈ cq ′ .
Hence (q|χ| ≤ 2 by (63) and |χ| ≪ 1),
(
)
∂x χ = ∂x |χ|eiη = eiη |χ|′ + iη ′ eiη |χ| = O(q ′ )
20
Theorem 5.3. For 0 < |χ| << 1 , we have that | cos η| ≈ 0 or
sin η = −2 cos η.
√1
5
and
Proof. We have that
∂x q(1 + O(χ)) = ∂x Imχ = ∂x (|χ| sin η) = |χ|′ sin η + |χ| cos η∂x η
= sin2 η∂x q(1 + O(χ)) − cos η(sin η + cos η + O(χ))∂x q
[
]
= sin2 η − cos η(sin η + cos η) + O(χ) ∂x q.
It follows that
sin2 η − cos η(sin η + cos η) + O(χ) = 1.
The solutions of this trigonometric identity (up to O(χ)), are cos η = 0, and
tan η = −2, from which the result of the theorem follows.
Remark As long as χ is small, the solution is trapped around one of the
above solutions of the trigonometric identity. The case of cos η = 0 corresponds to essentially no mass change of the soliton energy; the other cases
corresponds to the soliton either gaining or losing, monotonically, energy
(and mass)
We can now prove the main bound on χ:
Theorem 5.4. Multiplicative Error Bound
δ
, and x is in the transition region,
For t ≤ ωε
|χ(x)| ≤ δ.
Proof. Equation (59) yields
i∂t ⟨F (x), χ⟩ =
∂x S
∂x χ⟩ + 2⟨F, SReχ⟩ + h.o.t.
S
Therefore, for F supported in [a, b] we have
∫ b
∫ b
2
|∂x F (x)|dx
∂x F (x)∂x χdx ≤ sup |∂x χ|
|⟨F (x), −∂x χ⟩| = |⟨∂x F, ∂x χ⟩| = ⟨F (x), −∂x2 χ⟩ + ⟨F, W (x, t)χ⟩ − 2⟨F (x),
x∈[a,b]
a
≤ C|q ′ |(b − a) sup |∂x F (x)| ≤ C|q ′ |
x∈[a,b]
21
a
by choosing

αx,





1,
F (x) =





−αx,
so that
∫
0≤x≤
b−a
4
≤x≤
3(b−a)
4
b−a
;
4
3(b−a)
;
4
≤x≤b
b
|∂x F (x)|dx ≤ C
a
does not depend on b − a.
The second term is higher order in the sense that
(
)
⟨F, W (x, t)χ⟩ = O e−1/ω ||F χ||.
The third term is bounded similarly
∫ b
⟨F (x), ∂x S ∂x χ⟩ ≤ C
|∂x χ| ≤ C|b − a|q ′ .
S
a
(
)
The rest of the terms are higher order or O e−1/ω χ
These estimates together with equation (59) imply that
|∂t ⟨F, χ⟩| ≤ C|b − a|q ′ + C1 q ′ .
So,
∫
|⟨F, χ⟩t | ≤ C|b − a|
t
′
∫
t
q (s)ds + C1
0
(67)
q ′ (s)ds
(68)
0
for t ≤ 2t0 , and recall that
v = q ′ (t) ∼ ε2 ω 2 t, for t ≤ t0 .
More generally for t ≤ t0 we have
∫ t
q ′ (s)ds ∼ ε2 ω 2 t2 /2, for t < t0 .
0
On an interval of size one we use the condition ∂x χ = O(q ′ ) ≪ 1 in order
to estimate
∫ a+1
∫ a+1
|χ|dx =
e−iη(x) χdx =
a
a
22
e
−iη(x)
∫
x
a
Then
∫ a+1
∫
≤
|χ|dx
a+1
a
a
a+1 ∫
χ(y)dy +
a+1
′
∫
a
a
∫
χ(y)dy +
a+1
a
y
iη (y)
χ(y ′ )dy ′ .
a
∫
|η (y)|dy sup a≤y≤a+1
′
a
y
χ(y )dy . (69)
′
′
Now, if at some point |χ(a)| ≤ Cq ′ , then it follows from (66) and (71)
below that
sup |χ| ≤ C|b − a|q ′ + C1 q ′ .
(70)
[a,b]
Therefore we only need to consider an interval [a, b] with
|χ(y)| > Cq ′ ⇒ |η ′ (y)| ≤
2q ′
≤ C < ∞.
|χ|
We have used that equation (66) implies
|χ(b)|2 − |χ(a)|2
≤ C|b − a|q ′ .
sup |χ|
x∈[a,b]
We make |χ(b)| = supx∈[a,b] |χ| by varying b, and as a result we get
|χ(a)|2
≤ C|b − a|q ′ .
sup |χ|
sup |χ| −
x∈[a,b]
(71)
x∈[a,b]
Then by 67
∫
a+1
|χ|dx ≤
a
∫
sup y∈[a,a+1]
y
a
∫ t
∫ t
′
χ(y )dy ≤ C|b − a|
q (s)ds + C1
q ′ (s)ds.
′
′
0
0
(72)
So, in particular,
{
′
∫
|χ(y)| ≤ max Cq + C1
t
}
′
q (s)ds
≤ Cε2 ω 2 t2 , for t ≤ t0 =
0
which means that |χ(x)| ≤ δ for all t ≤ δ/(ωε).
23
1
ωε
(73)
6
Momentum decay
In this section we prove that if the radiation is outgoing from the boundary
of the well, and if the soliton energy is monotonic decreasing (Ė ≤ 0), then
the soliton slows down, monotonically. To this end, we modify the Ansatz
so as to include the slow time dependent change in the oscillation, explicitly
in the ∫soliton term. That is we replace S(x − X) by S(x − a(t)), with
t
a(t) ≡ 0 v(s)ds + D(t).
Theorem 6.1. Assume that for the NLS as before, we have that ∂x q > 0,
that χ is small, and that Ė < 0. Then the velocity of the center of the soliton,
v(t), is monotonic decreasing in time, according to the Dissipation equation
below.
Proof. Let
∂ψ
= (−∂x2 + V )ψ + λ|ψ|2 ψ
∂t
−ESE = (−∂x2 + V )SE + λSE3
i
where λ < 0. The function S of the form
SE (x, t) = eiθh (x,t) SE (x − a(t))
solves NLS equation if (V = Vh )
∫ t
θh =
E(s)ds + γ(t) + v(t)x/2.
0
Momentum identity
(n)
Let FK = FK (|x| ≥ K) and such that |FK | . K −n . Then
[
]
∂V
ψ⟩+⟨ψ, FK λ(−∂x |ψ|2 )FK + ı[−∂x2 , FK pFK ] ψ⟩
∂x
∫
λ
2 ∂V
2
= −⟨ψ, FK
ψ⟩ + ⟨ψ, ı[−∂x , FK pFK ]ψ⟩ +
∂x FK FK |ψ|4 dx
∂x
2
∂t ⟨ψ, FK pFK ψ⟩ = −⟨ψ, FK2
where
p = −i∂x .
Ansatz:
[
]
ψ = eiθh (x,t) SE(t) (x − a(t)) + R(x, t)
24
with orthogonality as before.
Then
i
∂R
= (−∂x2 + Vt − E(t))R + 2λS 2 R + λS 2 R
∂t
+λ|R|2 R + O(SR2 ) + O(Ė, γ̇)R + (V − Vt + xv̇/2)(R + S).
(74)
Vt = V (x − a(t)).
Now we use the Ansatz in the momentum identity
∂t ⟨eiθ S, FK pFK eiθ S⟩ = ∂t ⟨S, FK (p + v/2)FK S⟩ = ∂t ⟨S, FK (v/2)FK S⟩ + 0
∫
λ
iθ
iθ
2 ∂V
2 ∂V
= −∂t ⟨e R, FK pFK e R⟩ − ⟨S, FK
S⟩ − ⟨R, FK
R⟩ +
FK′ |ψ|4 dx
∂x
∂x
2
[
]
∂V
− ∂t ⟨eiθ S, FK pFK eiθ R⟩ + c.c. − 2Re⟨S, FK2
R⟩ + ⟨ψ, ı[−∂x2 , MK ]ψ⟩. (75)
∂x
⟨ψ, ı[−∂x2 , MK ]ψ⟩ = ⟨eıθ S, ı[−∂x2 , MK ]eıθ S⟩
+⟨eıθ R, ı[−∂x2 , MK ]eıθ R⟩ + 2Re⟨eıθ S, ı[−∂x2 , MK ]eıθ R⟩.
(76)
Estimates:
Define
MK ≡ FK pFK .
∂t ⟨S, FK2 (v/2)S⟩ − ⟨eıθ S, ı[−∂x2 + V + λS 2 , MK ]eıθ S⟩
= (v̇/2)⟨S, FK2 S⟩+(v/2)∂t ⟨S, FK2 S⟩−⟨S, ı[(p+v/2)2 +V +λS 2 , FK (p+v/2)FK ]S⟩
= (v̇/2)⟨S, FK2 S⟩ + T 1 − ⟨S, ı[V − Vt + ω 2 ax, MK ]S⟩ + ⟨S, FK2 S⟩ω 2 a(t)
−⟨S, ı[p2 + Vt + λS 2 , FK (p + v/2)FK ]S⟩ − ⟨S, ı[pv, FK (p + v/2)FK ]S⟩
= (v̇/2)⟨S, FK2 S⟩ + T 1 − v⟨S, [FK′ pFK + FK pFK′ ]S⟩ + ⟨S, FK2 S⟩ω 2 a(t)
2
S⟩ − v 2 ⟨S, ı[p, FK2 ]S⟩ = ∂t [v/2⟨S, FK2 S⟩]+
+O(S 4 ) + ⟨S, sn (x)V(3)
O(S 4 ) + O(S 2 )O(v 2 + a) + ω 2 a(t)⟨S, FK2 S⟩ + ⟨S, sn (x)V32 S⟩.
Here sn (x) is the sign of x. We used that
2
FK ı[p, V − Vt + ω 2 a]FK ≡ sn (x)V(3)
FK2 .
25
(77)
Here, we applied the Virial Theorem for the commutator expectation on S.
The v 2 term drops since it is the integral of a product of the symmetric
function S 2 , (up to corrections of order a) with the antisymmetric function
(FK2 )′ , the last dropped since the real part of the expectation of p , on real
functions , vanishes.
−∂t ⟨eiθ R, FK pFK eiθ R⟩ + ⟨eıθ R, ı[−∂x2 + V, MK ]eıθ R⟩
= −∂t ⟨R, (v(t)/2)FK2 R⟩ − ∂t ⟨R, FK pFK R⟩ + ⟨eıθ R, ı[−∂x2 + V, MK ]eıθ R⟩
= −(v̇/2)⟨R, FK2 R⟩ − (v/2)∂t ⟨R, FK2 R⟩ − ∂t ⟨R, FK pFK R⟩
+⟨R, ı[(p + v/2)2 + V, FK (p + v/2)FK ]R⟩
= −(v̇/2)⟨R, FK2 R⟩ − (v/2)∂t ⟨R, FK2 R⟩ − ⟨R, ı[−∂x2 + Vt − E + 2λS 2 , MK ]R⟩
−⟨R, ı[V −Vt +xv/2, MK ]R⟩−⟨(V −Vt +xv/2)S, MK R⟩−⟨R, MK (V −Vt +xv/2)S⟩
[
]
2
2
2
+⟨R ı[p + V, MK + (v/2)FK ] + ı[pv, FK (p + v/2)FK ] + O(SR + S C, Ė, γ̇)MK R⟩
[
]
= −(v/2)∂t ⟨R, FK2 R⟩+v⟨R, ı[p2 , (FK2 /2)] + (∂x FK (p + v/2)FK + FK (p + v/2)∂x FK ) R⟩
+⟨R, O(S 2 , SR, O(Ė, γ̇))MK R⟩−⟨(V −Vt +xv/2)S, MK R⟩−⟨R, MK (V −Vt +xv/2)S⟩
[
]
= −(v/2)∂t ⟨R, FK2 R⟩ + v⟨R, ı[p2 , (FK2 /2)] + (∂x FK pFK + FK p∂x FK ) R⟩
+v 2 ⟨R, ∂x FK FK R⟩+h.o.t.−⟨(V −Vt +xv/2)S, MK R⟩−⟨R, MK (V −Vt +xv/2)S⟩
= −(v/2)∂t ⟨R, FK2 R⟩ + (v 2 /2)⟨R, ∂x FK FK R⟩ + (v 2 /4)⟨R, FK ∂x FK R⟩ + h.o.t
+ı⟨R, [p2 , FK2 ]R⟩v − 2Re⟨Ṽ S, pR⟩.
(78)
Here Ṽ ≡ (V − Vt + xv/2). We also chose FK Ṽ = Ṽ , ∂x FK Ṽ = 0. We have
used in the above the following:
∂x FK pFK + FK p∂x FK = ı[p, FK pFK ] = ı[p2 , FK2 ]/2
T1
∫
∫
∂S 2
∂S
= v(t)Ė
FK Sdx − v(t)ȧ
SFK2 dx
∂E
∂x
∫
∫
∂S 2
= v(t)Ė
F Sdx + v(t)ȧ FK S 2 ∂x FK dx
∂E K
(v(t)/2)∂t ⟨S, FK2 S⟩
26
∫
= v(t)Ė
∫
∂S 2
F Sdx + (v(t)/2)ȧ FK [−S(x)2 + S(x − a)2 ]∂x FK dx
∂E K
∫
+(v(t)/2)ȧ FK S(x)2 ∂x FK dx(= 0)
∫
∫
∂S 2
∂S(x)
= v(t)Ė
FK Sdx − v(t)ȧa(1 + o(a)) FK S
∂x FK dx
(79)
∂E
∂x
Finally, we derive the statement of the theorem and the Dissipation Equation:
Dissipation Equation
]
d [
(v/2)⟨S, FK2 S⟩
dt
= −(v/2)∂t ⟨R, FK2 R⟩ − ω 2 a(t)⟨S, FK2 S⟩ + v 2 [⟨R, FK ∂x FK R⟩]
2
2
+⟨ψ, sn (x)V(3)
ψ⟩ − ⟨R, sn (x)V(3)
R⟩ + h.o.t.
(80)
We used that due to antisymmetry of ∂x FK , ⟨S, FK ∂x FK S⟩ = O(a).
Cross Terms
[
]
2Re⟨eıθ S, ı[−∂x2 , MK ]eıθ R⟩ − ∂t ⟨eıθ S, MK eıθ R⟩ + c.c.
= 2Re⟨S, ı[p2 + pv, MK ]R⟩ − ∂t [⟨S, FK (p + v/2)FK R⟩ + c.c.]
[
]
= 2Re⟨S, ı[p2 + pv, MK ]R⟩ − ⟨∂E S Ė − ∂x S ȧ, FK (p + v/2)FK R⟩ + c.c.
[
]
+ v̇⟨S, FK2 R⟩ + c.c. + ⟨S, FK (∂x )FK (−2∂t ImR)⟩ − (v/2)⟨S, FK2 2ReṘ⟩
[
]
= v̇⟨S, FK2 R⟩ + c.c. + ȧ⟨∂x S, FK2 ReR⟩(v/2) − Ėv⟨∂E S, FK2 ReR⟩
[
]
+Re ȧ⟨∂x S, MK R⟩ − Ė⟨∂E S, MK R⟩ .
(81)
We used above the following observations:
Term 2
For f, g real valued,
⟨S, FK pFK S⟩ = 0, Re⟨f, pg⟩ = 0,
since SFK is real function.
Term 3
27
∂V
⟨S, FK2
∂x
∫
S⟩ =
S
∫
2 ∂V
∂x
FK2 dx
[S 2 (x − a) − S 2 (x)]
∫
= −2a(1 + o(a))
∫
where we have used
=
S 2 (x)
S(x)
∂V 2
F dx
∂x K
∂S(x) ∂V 2
F dx
∂x ∂x K
∂V
FK dx = 0
∂x
if V (x) = V (−x).
Term 4
Re⟨Ṽ S, pR⟩ = ⟨Ṽ S, ∂x ImR⟩ ∼ ⟨S 2 Ṽ ′ /2, Imχ⟩
(
)
(
)
= O ∂x S(K)S(K)ω 2 (|Imχ| + |vq |) a(t) = O ∂x S(K)S(K)ω 2 (ϵ2 ω 2 t + δ) ,
(82)
δ
for |t| ≤ ϵω
.
Term 5 (cross term)
⟨S, FK ∂x FK ∂t SImχ⟩ = ⟨FK S, ∂t FK S∂x Imχ⟩ + ⟨FK S, ∂t ∂x (FK S)Imχ⟩
= ⟨∂t (FK S)2 /2, vq ⟩ + ⟨(FK S)2 , v̇q ⟩+
⟨(FK S)(∂t ∂x (FK S)Imχ)⟩ − ⟨(FK S)2x /2, ∂t ∂x (Imχ)⟩
(
)
2
2 2
+⟨(FK S)x /2, Imχ⟩ = O ϵ ω S∂x S(K)[(ȧ + Ė)t + 1] a(t).
(83)
term 6 (cross term)
v⟨S, FK2 ReṘ⟩ = v⟨S, FK2 HImχS⟩ − v⟨S, FK2 Ṽ S⟩
using that
v⟨S, [FK2 , −∂x2 ]SImχ⟩ − v⟨S, FK2 Ṽ S⟩
1
= O( )vS 2 (K)vq + O(v)aS 2 (K) = o(1)vS 2 (K),
K
∫
Ṽ = 0
(84)
on support of the function FK , and FK2 S 2 is symmetric up to corrections of
order a.
28
7
Calculation of phases
We analyze here equations (42) and (43):
1 + |χ| cos η = |1 + χ| cos q
|χ| sin η = |1 + χ| sin q
(85)
(86)
We need the derivative
∂x |1 + χ| =
=
1
[|χ||χ|′ + ∂x Reχ]
|1 + χ|
1
[|χ||χ|′ + |χ|′ cos η − |χ| sin η ′ ] ,
|1 + χ|
and
1−
1
|1 + χ| − 1
|1 + χ|2 − 1
2Reχ + |χ|2
=
=
=
|1 + χ|
|1 + χ|
|1 + χ|[1 + |1 + χ|]
|1 + χ|[1 + |1 + χ|]
= [Reχ + |χ|2 /2][1 − 3Reχ + O|χ|2 ] = |χ| cos η + |χ|2 /2 − 3|χ|2 cos2 η + O(|χ|3 )
Now differentiate equations (42) and (43) with respect to x and get
|χ|′ cos η − η ′ |χ| sin η = |1 + χ|′ cos q − |1 + χ| sin qq ′
|χ|′ sin η + η ′ |χ| cos η = |1 + χ|′ sin q + |1 + χ| cos qq ′
Then
(87)
(88)
[
]
[
]
cos q + |χ|/ cos η
cos q
′
|χ| cos η 1 −
− η |χ| sin η 1 −
|1 + χ|
|1 + χ|
[
]
cos q
′
′
= −|1 + χ|q sin q|χ| sin η − (cos η + |χ|)
+
|1 + χ|
[
]
sin q
′
η |χ| cos η + sin η
= |1 + χ|q ′ cos q
|1 + χ|
′
Now we calculate the determinant of this equation
[
]
[
]
cos q+|χ|/ cos η
cos q
;
−|χ| sin η 1 − |1+χ|
cos η 1 −
|1+χ|
]
[
]
Det [
cos q
cos q
; |χ| cos η + sin η |1+χ|
sin η − (cos η + |χ|) |1+χ|
29
=
(89)
[
]
|χ|2 cos q
cos q
−(sin η + cos η)
+ |χ| 1 −
·
|1 + χ|
|1 + χ|
[
]
]]
[
[
cos q
cos q
cos η cos η + sin η
+ sin η sin η − cos η
|1 + χ|
|1 + χ|
[
]
|χ|2 cos q
cos q
= −(sin η + cos η)
+ |χ| 1 −
|1 + χ|
|1 + χ|
[
( 2 )]
q
= −|χ|2 sin η 1 + O
− |χ|3 cos2 η.
|χ|
In order to find |χ|′ we need the determinant
[
]
cos q
′
−|1
+
χ|
sin
qq
;
−|χ|
sin
η
1
−
|1+χ|
[
]
Det cos q
′
|1 + χ| cos qq ; |χ| cos η + sin η |1+χ|
[
(90)
=
[
]
[
]
]
cos q
cos q
q |χ||1 + χ| − sin q cos η + sin η
+ sin η 1 −
cos q =
|1 + χ|
|1 + χ|
′
q ′ |χ||1 + χ| [− sin q(cos η + sin η) + |χ| sin η cos η]
+O(q ′ |χ|3 sin η(sin η + cos η)).
Then
|χ|′ =
(91)
q′
|1 + χ| [sin q(1 + cot η) − |χ| cos η]
|χ|
In order to find η ′ we need the determinant
]
[
cos q+|χ|/ cos η
;
−|1 + χ| sin qq ′
cos η 1 −
|1+χ|
[
]
Det cos q
; |1 + χ| cos qq ′
sin η − (cos η + |χ|) |1+χ|
=
−q ′ |χ| cos q(sin q + cos q)+
]
[
]]
[
[
cos q
cos q
′
q |1 + χ| cos q cos η 1 −
+ sin q sin η − cos η
≈
|1 + χ|
|1 + χ|
[
]
−q ′ |χ| + q ′ |1 + χ| |χ| cos2 η + q [sin η − cos η] ≈
−q ′ |1 + χ||χ| sin η[sin η + cos η](1 + O(|χ|)).
Then
η′ = −
q′
[cos η + sin η + O(|χ|)]
|χ|
30
(92)
(93)
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....
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32